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\begin{document}

\input{213-Lecture-Macros}		%% 	macros, settings, et.c


\title{Math 213 - Equations of Lines and Planes, II}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{September 9, 2019}
\maketitle


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%	
%		Due Date Reminders
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Reminders}

\begin{itemize}

\item	Access your WebWork account \emph{only through Canvas}!

\medskip

\item	Homework A3 on section 12.5 is due Wednesday
\item	Homework A4 on section 12.6 is due Friday
\item	Quiz \# 2 on sections 12.3-12.4 takes place in Thursday recitation

\medskip
		
	
\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	Unit Outline
%
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\begin{frame}
\frametitle{Unit I: Geometry and Motion in Space}
\begin{itemize}

\item[{12.1}]		Lecture 1: Three-Dimensional Coordinate Systems
\item[{12.2}] 		Lecture 2: Vectors in the Plane and in Space
\item[{12.3}]		Lecture 3:The Dot Product
\item[{12.4}]		Lecture 4:The Cross Product
\item[{12.5}]		Lecture 5:	Equations of Lines and Planes, I
\begingroup
\color{red}
\item[{12.5}]		Lecture 6: Equations of  Lines and Planes, II
\endgroup
\item[{12.6}]		Lecture 7: Surfaces in Space
\item[{13.1}]		Lecture 8: Vector Functions and Space Curves
\item[{13.2}]		Lecture 9 Derivatives and Integrals of Vector Functions
\item[{}]				Lecture 10:  Exam I Review
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	Learning Goals
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Learning Goals}
\begin{itemize}
\item		Review dot, cross, and triple scalar products
\item		Review equations of lines and planes
\item		Sketch and visualize lines and planes
\item		Learn how find the distance from a point to a plane
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%		New material begins here...
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\begingroup
\small

\section{Dot, Cross, Triple}

\begin{frame}
\frametitle{Dot Product, Cross Product, Triple Product}
\footnotesize

\medskip

\begin{center}
{\renewcommand{\arraystretch}{1.5}
\begin{tabular}{lllll}
	&		\textbf{Formula} 	&	\textbf{Type}	&	\textbf{Geometry}	&	\textbf{Zero if$\ldots$}	\\
\\
\textbf{Dot}	&	$\bfa \cdot \bfb$		&Scalar	&	Projections	&	$\bfa$, $\bfb$ orthogonal\\
\\
\textbf{Cross}	&	$\bfa \times \bfb$	&Vector	&	Area of a	&	$\bfa$, $\bfb$ parallel	\\
&&& Parallelogram	\\
\\
\textbf{Triple}	&	$\bfa \cdot (\bfb \times \bfc)$	& Scalar	& Volume of a &	$\bfa, \bfb, \bfc$ coplanar \\
&&&	Parallelepiped	\\
\end{tabular}
}
\end{center}

\end{frame}

\section{Lines and Planes}

\begin{frame}
\frametitle{Lines and Planes}
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.45]
\drawxyz{5}
\draw		[thick,dashed,cyan]		(0,1,1)	--	++(4,4,4);
\uncover<2->
{
\draw[black,fill=black]	(1,2,2)	circle(0.1cm);
\node[left]	at	(1,2,2)	{\scriptsize{$(x_0,y_0,z_0)$}};
}
\uncover<3->
{
\draw[thick,red,->,>=stealth]	(1,2,2)	--	(2,3,3);
\node[right] at (2,3,3)		{\scriptsize{$\langle a,b,c \rangle$}};
}
\end{tikzpicture}
\begin{tikzpicture}[scale=0.45]
\drawxyz{5}
\draw[dashed,blue,fill=blue!30,opacity=0.5]	(0,0,3)	--	(3,0,0)	--	(0,3,0)	--	(0,0,3);
\uncover<4->
{
\draw[black,fill=black]	(1,1,1)	circle(0.1cm);
\node[right]	at	(1,1,1)	{\scriptsize{$(x_0,y_0,z_0)$}};
}
\uncover<5->
{
\draw[red,thick,->,>=stealth]	(1,1,1)	--	(3,3,3);
\node[right] at (3,3,3)		{\scriptsize{$\langle a,b,c \rangle$}};
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
To specify the equation of a \textbf{line} $L$, you need:
\begin{itemize}
\uncover<2->
{
\item	A point $(x_0,y_0,z_0)$ on $L$\
}
\uncover<3->
{
\item	A vector $\langle a,b,c \rangle$ in the direction of $L$
}
\end{itemize}

\bigskip
To specify the equation of a \textbf{plane}, you need:
\begin{itemize}
\uncover<4->
{
\item	A point $(x_0, y_0, z_0)$ on the plane
}
\uncover<5->
{
\item	A vector $\bfn = \langle a,b,c \rangle$ normal to the plane
}
\end{itemize}
\end{minipage}

\end{frame}

\begin{frame}
\frametitle{Hot Tip - Planes Made Simple}
\begin{minipage}{0.40\textwidth}
\begin{tikzpicture}[scale=0.40]
\drawxyz{5}
\draw[dashed,blue,fill=blue!30,opacity=0.5]	(0,0,3)	--	(3,0,0)	--	(0,3,0)	--	(0,0,3);
{
\draw[black,fill=black]	(1,1,1)	circle(0.1cm);
\node[right]	at	(1,1,1)	{\scriptsize{$(x_0,y_0,z_0)$}};
}
{
\draw[red,thick,->,>=stealth]	(1,1,1)	--	(3,3,3);
\node[right] at (3,3,3)		{\scriptsize{$\langle a,b,c \rangle$}};
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.5\textwidth}
\footnotesize
The equation of a plane is
$$ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 $$
or
$$ ax + by + cz = d $$

\medskip
\begin{itemize}
\item[Step 1.]		Determine $\langle a,b,c \rangle$ from geometry
\item[Step 2.]		Find $d$ by substituting in $x_0, y_0, z_0$
\end{itemize}

\smallskip

\hrulefill

\medskip

Example: Find the equation of a plane parallel to the plane 
$$x-y+2z=0$$
through the point $(2,2,2)$. 

\end{minipage}

\end{frame}


\begin{frame}
\frametitle{Hot Tip - Planes Made Simple}
\begin{minipage}{0.40\textwidth}
\begin{tikzpicture}[scale=0.40]
\drawxyz{5}
\draw[dashed,blue,fill=blue!30,opacity=0.5]	(0,0,3)	--	(3,0,0)	--	(0,3,0)	--	(0,0,3);
{
\draw[black,fill=black]	(1,1,1)	circle(0.1cm);
\node[right]	at	(1,1,1)	{\scriptsize{$(x_0,y_0,z_0)$}};
}
{
\draw[red,thick,->,>=stealth]	(1,1,1)	--	(3,3,3);
\node[right] at (3,3,3)		{\scriptsize{$\langle a,b,c \rangle$}};
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.5\textwidth}
\footnotesize
The equation of a plane is
$$ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 $$
or
$$ ax + by + cz = d $$

\medskip
\begin{itemize}
\item[Step 1.]		Determine $\langle a,b,c \rangle$ from geometry
\item[Step 2.]		Find $d$ by substituting in $x_0, y_0, z_0$
\end{itemize}


\hrulefill

\medskip

Example: Find the equation of a plane orthogonal to the line 
$$ (x,y,z) = (-7,0,0) + t (-7,3,3)$$
which passes through the point $(0,0,-7)$. Give your answer 
in the form $ax+by+cz=d$ where $a=7$. 

\end{minipage}

\end{frame}

\begin{frame}
\frametitle{Hot Tip - Sketching Planes Made Simple}
\begin{minipage}{0.40\textwidth}
\begin{tikzpicture}[scale=0.50]
\drawxyz{5}
\draw[dashed,blue,fill=blue!30,opacity=0.5]	(0,0,3)	--	(3,0,0)	--	(0,3,0)	--	(0,0,3);
{
\draw[red,thick,->,>=stealth]	(1,1,1)	--	(3,3,3);
\node[right] at (3,3,3)		{\scriptsize{$\langle a,b,c \rangle$}};
}
\uncover<2->
{
\node[right]	at	(0,-0.4,3)		{\scriptsize{$(d/a,0,0)$}};
\node[above] 	at (4,0,0)			{\scriptsize{$(0,d/b,0)$}};
\node[right]	 at	 (0,3,0)			{\scriptsize{$(0,0,d/c)$}};
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.5\textwidth}
\footnotesize
The equation of a plane is
$$ ax + by + cz = d $$

\uncover<2->
{
\medskip
To sketch the plane with this equation, you can find
the $x$-, $y$-, and $z$-intercepts from the equation:

$$x=d/a, \quad y=d/b, \quad z=d/c $$
}

\hrulefill

\medskip

\uncover<3->
{
Sketch the part of the plane $$2x+y+3z=4$$ in the first octant and label the $x$- , $y$-, and $z$-intercepts.
}
\end{minipage}

\end{frame}

\section{Visualizing}

\begin{frame}
\frametitle{Intersecting, Parallel, and Skew Lines}
\begin{minipage}{0.45\textwidth}
\begin{minipage}{0.9\textwidth}
\begin{tikzpicture}[scale=0.4]
\uncover<2->
{
\draw[thick,->,>=stealth]	(0,0)		--	(4,0);
\draw[thick,->,>=stealth]	(0,0)		--	(0,5);
\node[right] 	at (4,0)		{\scriptsize{$x$}};
\node[above]	at	(0,5)		{\scriptsize{$y$}};
}
\only<2>
{
\draw[cyan,thick]	(-2,2)		--	(2,5);
\draw[cyan,thick]	(-1,2)		--	(3,5);
}
%%
\uncover<3->
{
\draw[cyan,thick]	(-1,1)		--	(1,5);
\draw[cyan,thick]	(-1,5)		--	(1,1);
}
%%
\end{tikzpicture}
\end{minipage}

\medskip

\begin{minipage}{0.9\textwidth}
\begin{tikzpicture}[scale=0.4]
\uncover<4->
{
\drawxyz{5}
}
%%
\only<5>
{
\draw[thick,cyan]	(0,0,4)	--	++(2,4,-3);
\draw[thick,cyan]	(0,0,0)	--	++	(2,4,-3);
}
%%
\only<6>
{
%\draw[thick,cyan]	(0,0,4)	--	++(2,1,-3);
\draw[thick,cyan]	(0,0,4)	--	++(6,3,0);
\draw[dashed]		(0,0,4)	--	++(6,0,0)	--	++(0,3,0);
%\draw[thick,cyan]	(0,2,0)	--	++(1,2,-1.5);
\draw[thick,cyan]	(0,2,0)	--	++(2,4,0);
}
%%
\only<7-8>
{
\draw[thick,cyan]			(0,2,0)	--	++(7,0,7);
\draw[thick,cyan]			(0,0,0)	--	++(7,7,7);
\draw[black,fill=black]		(2,2,2)	circle(0.1cm);
\draw[dashed]				(0,0,0)	--	(7,0,7)	--	(7,7,7);
}
\end{tikzpicture}
\end{minipage}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
In two-dimensional space, two lines $L_1$  and $L_2$ can be 
\begin{itemize}
\uncover<2->
{
\item	\emph{parallel}, or
}
\uncover<3->
{
\item	\emph{intersecting}
}
\end{itemize}

\bigskip

\uncover<4->
{
In three dimensions, two lines $L_1$ and $L_2$ can be 
}
\begin{itemize}
\uncover<5->
{
\item		\emph{parallel},
}
\uncover<6->
{
\item		\emph{skew}, or 
}
\uncover<7->
{
\item		\emph{intersecting}
}
\end{itemize}

\bigskip

\uncover<8->
{How do you tell which is which?}
\end{minipage}
\end{frame}

\begin{frame}
\frametitle{Intersecting, Parallel, and Skew Lines}

\footnotesize
$$ \bfr(t) = \bfr_0(t) + t \bfv $$
\begin{itemize}
\item	Two lines are parallel if the corresponding vectors $\bfv$ are parallel
\item	If not parallel, two lines intersect if we can solve for the point of intersection
\item	If not parallel, and nonintersecting, they are skew
\end{itemize}


\hrulefill

\medskip
Determine whether the following pairs of lines are parallel, intersect, or are skew. If they intersect, find the points of intersection.

\begin{enumerate}
\item  	$L_1: 	x=2+s, \quad 	y=3-2s, \quad 	z=1-3s$ \\[5pt]
			$L_2: 	x=3+t, \quad  y = -4+3t, 	\quad	z=2-7t$
		
\smallskip

\item	$L_1:	\dfrac{x}{1} = \dfrac{y-1}{-1} = \dfrac{z-1}{-3}$, \hphantom{XXXX}
		$L_2:	\dfrac{x-2}{2}	=	\dfrac{y-3}{-2}	=	\dfrac{z}{7} $
\end{enumerate}

\end{frame}

\begin{frame}
\frametitle{Intersecting and Parallel Planes}
\begin{minipage}{0.45\textwidth}
\begin{center}

\begin{minipage}{0.9\textwidth}
\begin{tikzpicture}[scale=0.4]
\uncover<2->
%% parallel planes
{
\drawxyz{5}
\shade[left color=blue!80,right color=blue!5,opacity=0.75]	(0,0,4)		--	++(4,0,0)	--	++(0,2,-4)
	--	++(-4,0,0)	--	++(0,-2,4);
\shade[left color=red!80,right color=red!10,opacity=0.75]			(0,2,4)		--	++(4,0,0)	--	++(0,2,-4)
	--	++(-4,0,0)	--	++(0,-2,4);
}
\end{tikzpicture}
\end{minipage}

\medskip

\begin{minipage}{0.9\textwidth}
\begin{tikzpicture}[scale=0.4]
\uncover<3->
%% intersecting planes and the line of intersecction
{
\drawxyz{5}
\shade[left color=blue!80,right color=blue!5,opacity=0.5]	(0,0,4)		--	++(4,0,0)	--	++(0,3,-6)
	--	++(-4,0,0)	--	++(0,-3,6);
\shade[left color=red!80,right color=red!10,opacity=0.5]		(0,5,4)		--	++(4,0,0)	--	++(0,-6,-4)
	--	++(-4,0,0)	--	++(0,6,4);
\draw[thick,dashed,cyan]		(0,1.25,1.5)	--	++(4,0,0);
}
\uncover<4->
%% two normal vectors and vector along the line
{
\draw[blue,thick,->,>=stealth]	(2,1.25,1.5)	--	++(0,3,1.5);
\draw[red,thick,->,>=stealth]	(2,1.25,1.5)	--	++(0,1.5,-2.25);
\draw[thick,cyan,->,>=stealth]	(2,1.25,1.5)	--	++	(2,0,0);
}
\end{tikzpicture}
\end{minipage}
\end{center}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
Two planes either
\begin{itemize}
\uncover<2->
{
\item		are parallel (if their normal vectors are parallel), or
}
\uncover<3->
{
\item		intersect in a line
}
\end{itemize}

\uncover<4->
{
A vector pointing along that line will be perpendicular to \emph{both} normal vectors
}

\smallskip

\hrulefill

\medskip

\uncover<5->
{
Find the line of intersection between the planes 
$$x+2y+3z=1$$ and $$x-y+z=1$$
}

\end{minipage}
\end{frame}

\section{Distances}

\begin{frame}
\frametitle{The Distance from a Point to a Plane}
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.5]
\shade[left color=blue!70,right color=blue!5]
	(0,0,0)	--	++	(0,0,5)	--	++(6,0,0)	--	++(0,0,-5)	--	++(-6,0,0)	--	++(0,0,5);
\draw[thick,black]						(1.5,0,4) --	++(1.5,0,-2);
\draw[dashed]							(3,0,2)	--	(3,3,2);
\draw[thick,red,->,>=stealth]		(3,0,2)	--	(3,2,2);
\draw[thick,cyan,->,>=stealth]		(1.5,0,4)	--	++(1.425,2.85,-1.9);
\draw[black,fill=black]					(3,3,2)	circle(0.1cm);
%%
\uncover<3->
{
\node[left] 		at (1.5,0,4)				{\scriptsize{$P_0$}};
}
\uncover<2->
{
\node[right] 	at	(3,3,2)				{\scriptsize{$P_1$}};
}
\uncover<3->
{
\node[left]		at	(3,1.5,2)			{\scriptsize{$\bfn$}};
}
\uncover<4->
{
\node[left] 		at (2.25,1.75,3)		{\scriptsize{$\bfb$}};
}
\draw[<->,>=stealth]	(3.75,0,1)	--	++(0,3,0);
\node[right] at (3.75,1.5,1)			{\scriptsize{$D$}};
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
To find the distance $D$
\uncover<2->{\emph{from} a point $P_1$}
\uncover<3->{\emph{to} a plane with normal vector $\bfn$ 
containing a point $P_0$:}

\medskip

\uncover<4->
{
Let $\bfb$ be the vector $\overrightarrow{P_0 P_1}$
}

\medskip

\uncover<5->
{
Then the distance $D$ is given by $\comp_{\bfn} \bfb$, or
$$
D = \frac{| \bfn \cdot \bfb|}{|\bfn|}
$$
}
\uncover<6->
{
If $$P_1 = P_1(x_1, y_1, z_1),$$ 
$$P_0=P_0(x_0,y_0,z_0),$$ 
then
$$
\bfb = 	(x_1-x_0,y_1-y_0,z_1-z_0)
$$
}
\end{minipage}

\end{frame}

\begin{frame}
\frametitle{The Distance from a Point to a Plane}
\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.5]
\shade[left color=blue!70,right color=blue!5]
	(0,0,0)	--	++	(0,0,5)	--	++(6,0,0)	--	++(0,0,-5)	--	++(-6,0,0)	--	++(0,0,5);
\draw[thick,black]	(1.5,0,4)	--	(3,0,2);
\draw[dashed]							(3,0,2)	--	(3,3,2);
\draw[thick,red,->,>=stealth]		(3,0,2)	--	(3,2,2);
\draw[thick,cyan,->,>=stealth]		(1.5,0,4)	--	++(1.425,2.85,-1.9);
\draw[black,fill=black]					(3,3,2)	circle(0.1cm);
%%
\node[left] 		at (1.5,0,4)				{\scriptsize{$P_0$}};

\node[right] 	at	(3,3,2)				{\scriptsize{$P_1$}};
\node[left]		at	(3,1.5,2)			{\scriptsize{$\bfn$}};
\node[left] 		at (2.25,1.75,3)		{\scriptsize{$\bfb$}};
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.55\textwidth}
\footnotesize
\begin{align*}
D 		&= 	\frac{| \bfn \cdot \bfb|}{|\bfn|}	\\
\bfb 	&= 	(x_1-x_0,y_1-y_0,z_1-z_0)	\\
\bfn	&= 	\langle a,b,c \rangle
\end{align*}

\medskip

If the plane's equation is
$$ ax + by + cz +d = 0$$
then
\begin{align*}
\bfn \cdot \bfb	&=	a(x_1-x_0) + b(y_1-y_0) + c(z_1-z_0)\\
					&=	a x_1 + by_1 +cz_1 + d
\end{align*}
so
$$ D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2+b^2 +c^2}}. $$

\end{minipage}
\end{frame}

\section{Summary}

\begin{frame}
\frametitle{Summary}
\footnotesize
\begin{itemize}
\item	We reviewed the basic facts about the dot product $\bfa \cdot \bfb$, the cross product $\bfa \times \bfb$, and the scalar triple product $\bfa \cdot (\bfb \times \bfc)$
\bigskip
\item	We reviewed how to write equations of lines and planes and $\ldots$
\medskip
\begin{itemize}
\item	How to determine whether two lines are parallel, perpendicular, or skew
\medskip
\item	How to determine whether planes are parallel or intersecting 
\end{itemize}
\bigskip
\item	We computed the distance from a point to a plane
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Homework}
\begin{itemize}
\item	Re-read section 12.5 and continue work on webwork A3 (due Wednesday)
\item Read and study section 12.6 on quadric surfaces
\item	Prepare for tomorrow's recitation on section 12.5
\end{itemize}
\end{frame}

\endgroup

\end{document}